Question

If $ x,\text{ }y,\text{ }z $ are positive then, minimum value of $ {{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}} $ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

We know, that $ AM\ge GM $
$ \therefore $ $ \frac{{{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}}{3}\ge $
$ \sqrt[3]{{{x}^{\log y-\log z}}.{{y}^{\log z-\log x}}.{{z}^{\log x-\log y}}} $ ..(i)
Now, $ \log \{{{x}^{\log y-\log z}}.{{y}^{\log z-\log x}}.{{z}^{\log x-\log y}}\} $
$ =\log \{{{x}^{\log y}}.{{x}^{-\log z}}.{{y}^{\log z}}.{{z}^{-\log x}}.{{z}^{\log x}}.{{z}^{-\log y}}\} $
$ =\log ({{x}^{\log y}})+\log ({{x}^{-\log z}})+\log ({{y}^{\log z}}) $
$ +\log ({{y}^{-\log x}})+\log ({{z}^{\log x}})+\log ({{z}^{-\log y}}) $
$ =\log y\log x-\log z\log x+\log z\log y $
$ -\log x\log y+\log x\log z-\log y\log z $
$ =0 $ $ \Rightarrow $ $ {{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}\ge 3 $
$ \therefore $ FromEq. (i),
$ {{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}\ge 3 $
Hence, minimum value is 3.